Can you chemical change or alter DNA

True or False?<br><br> Heat energy moves from cold areas to warm areas.

hichkok12 [17]

True because heat energy moves to the warmer areas.

4 0

2 years ago

An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth

yanalaym [24]

Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

Explanation:

The magnitude of the electric potential difference is the induced emf and is given by the equation:

$emf = l (v \times B)$

where l = length

v = velocity

B = magnetic field

As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. $v \times B = vB sin90 = vB\\$ ,

the induced emf = vlB, and thus increases.

The magnitude of the electric field potential difference between the wingtips increases

3 0

2 years ago

Need Some Help Please :)

joja [24]

1. The amount of energy carried by the wave is related to the Amplitude of the wave.
2. A mechanical wave requires an initial energy input, Once this initial energy is added the wave travels through the medium until all it's energy is transferred.

6 0

2 years ago

Read 2 more answers

List and Explain three ways study groups benefit your learning

vodomira [7]

Answer:

The answer

Explanation:

Thinking together, Better friendship, Makes teacher happy.

4 0

2 years ago

Read 2 more answers

A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse

kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball = 128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0

2 years ago